這裡主要補充本人遺漏的線段樹模板,沒有說明。
掃瞄線,並面積
const
int n =
1e5+5;
struct node
node (
double a,
double b,
double c,
int d):x
(a),
y1(b),y2
(c),
w(d)
}su[n]
;struct node
tr[n]
;double lisan[n]
;bool
cmp(node a, node b)
bool
_cmp
(double a,
double b)
int mx;
void
build
(int p,
int l,
int r)
void
change
(int p,
int l,
int r,
int w)
int mid =
(tr[p]
.l +tr[p]
.r)>>1;
if(l <= mid)
change
(p<<
1, l, r, w);if
(r > mid)
change
(p<<1|
1, l, r, w);if
(tr[p]
.lu >
0) tr[p]
.sum = lisan[tr[p]
.r+1
]- lisan[tr[p]
.l];
else tr[p]
.sum = tr[p<<1]
.sum + tr[p<<1|
1].sum;
}int
main()
sort
(lisan, lisan+_cnt)
;int m =
unique
(lisan, lisan+_cnt, _cmp)
- lisan;
sort
(su, su+cnt, cmp)
;build(1
,0, _cnt-1)
;double ans =
0.0;
for(
int i =
0; i < cnt-
1; i++
)printf
("test case #%d\n"
, t0++);
printf
("total explored area: %.2f\n\n"
, ans);}
}
掃瞄線,兩次以上交面積
const
int n =
1010
;struct node
node (
double a,
double b,
double c,
int d):x
(a),
y1(b),y2
(c),
w(d)
}su[n<<3]
;struct node
tr[n<<3]
;double lisan[n<<3]
;bool
cmp(node a, node b)
int mx;
void
build
(int p,
int l,
int r)
voidph(
int p)
void
change
(int p,
int l,
int r,
int w)
int mid =
(tr[p]
.l +tr[p]
.r)>>1;
if(l <= mid)
change
(p<<
1, l, r, w);if
(r > mid)
change
(p<<1|
1, l, r, w);ph
(p);
}void
print()
intmain()
sort
(lisan, lisan+_cnt)
;int m =1;
for(
int i =
1; i < _cnt; i ++)if
(lisan[i]
!= lisan[i-1]
) lisan[m++
]= lisan[i]
;// cout << m (su, su+cnt, cmp)
;build(1
,0, m-1)
;double ans =
0.0;
for(
int i =
0; i < cnt-
1; i++
)printf
("%.2f\n"
, ans +
0.000003);
}return0;
}
樹拉取dfs序列
void
dfs(
int x,
int fa)
r[x]
= cnt;
}void
build
(int p,
int l,
int r)
//......
}
序列單次覆蓋權值
序列中,每個元素有乙個權值,有m個覆蓋區間[l, r],k個減少覆蓋的區間[l, r]。乙個區間被覆蓋一次或以上就會貢獻一遍這個權值。否則貢獻值為0。有q次穿插詢問:問現在q這個序列的總權值。
參見掃瞄線,並面積。這裡要注意,不能查詢區間的權值資訊,只能查全序列的權值資訊。
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